Internet Surfer 2.0/Internet Surfer 2.0 (Wayzata Technology) (1996).iso/pc/text/mac/faqs.470

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Frequently Asked Questions (FAQS);faqs.470



You can demonstrate this with a tray, which you hold in your right hand
with the arm lowered, then rotate twice as you raise your arm and end
up with the tray above your head, rotated twice about its vertical
axis, but without having twisted your arm.

Also, by attaching strings to a sphere, it is possible to see that a
360 degree rotation will entangle the strings, which another 360 degree
rotation will disentangle.

Hospitals have machines which take out your blood, centrifuge it to take out
certain parts, then return it to your veins. Because of AIDS they must never
let your blood touch the inside of the machine which has touched others'
blood. So the inside is lined with a single piece of disposable branched
plastic tubing. This tube must rotate rapidly in the centrifuge where
several branches come out. Thus the tube should twist and tangle up the
branches. But the machine untwists the branches as in the above discussion.
At several hundred rounds per minute!

References
    P. A. M. Dirac's "scissors demonstration"
    R. Penrose and W.  Rindler
    Spinors and Space-time, vol. 1, p. 43
    Cambridge University Press, 1984,

    R. Feynman and S. Weinberg
    Elementary Particles and the Laws of Physics, p. 29
    Cambridge University Press, 1987

==> geometry/smuggler.p <==
Somewhere on the high sees smuggler S is attempting, without much
luck, to outspeed coast guard G, whose boat can go faster than S's. G
is one mile east of S when a heavy fog descends. It's so heavy that
nobody can see or hear anything further than a few feet. Immediately
after the fog descends, S changes course and attempts to escape at
constant speed under a new, fixed course. Meanwhile, G has lost track
of S. But G happens to know S's speed, that it is constant, and that S
is sticking to some fixed heading, unknown to G.

How does G catch S?

G may change course and speed at will. He knows his own speed and
course at all times. There is no wind, G does not have radio or radar,
there is enough space for maneuvering, etc.

==> geometry/smuggler.s <==
One way G can catch S is as follows (it is not the fastest way).

G waits until he knows that S has traveled for one mile. At that time, both
S and G are somewhere on a circle with radius one mile, and with its center
at the original position of S. G then begins to travel with a velocity that
has a radially outward component equal to that of S, and with a tangential
component as large as possible, given G's own limitation of total speed. By
doing so, G and S will always both be on an identical circle having its
center at the original position of S. Because G has a tangential component
whereas S does not, G will always catch S (actually, this is not proven
until you solve the o.d.e. associated with the problem).

If G can go at 40 mph and S goes at 20 mph, you can work out that it will
take G at most 1h 49m 52s to catch S.  On average, G will catch S in:

( -2pi + sqrt(3) ( exp(2pi/sqrt(3)) - 1 )) / 40pi    hours,

which is, 27 min and 17 sec.

==> geometry/table.in.corner.p <==
Put a round table into a (perpendicular) corner so that the table top
touches both walls and the feet are firmly on the ground.  If there is
a point on the perimeter of the table, in the quarter circle between
the two points of contact, which is 10 cm from one wall and 5 cm from
the other, what's the diameter of the table?

==> geometry/table.in.corner.s <==
Consider the +X axis and the +Y axis to be the corner.  The table has
radius r which puts the center of the circle at (r,r) and makes the
circle tangent to both axis.  The equation of the circle (table's
perimeter) is

    (x-r)^2 + (y-r)^2 = r^2 .

This leads to

     r^2 - 2(x+y) + x^2 + y^2 = 0

Using x = 10, y = 5 we get the solutions 25 and 5.  The former is the
radius of the table.  It's diameter is 50 cm.

The latter number is the radius of a table that has a point which
satisfies the conditions but is on the outside edge of the table.

==> geometry/tesseract.p <==
If you suspend a cube by one corner and slice it in half with a
horizontal plane through its centre of gravity, the section face is a
hexagon.  Now suspend a tesseract (a four dimensional hypercube) by one
corner and slice it in half with a hyper-horizontal hyperplane through
its centre of hypergravity.  What is the shape of the section
hyper-face?

==> geometry/tesseract.s <==
The 4-cube is the set of all points in [-1,1]^4 .
The hyperplane { (x,y,z,w) : x + y + z + w = 0 } cuts the 4-cube
in the desired manner.

Now, { (.5,.5,-.5,-.5), (.5,-.5,.5,-.5), (.5,-.5,-.5,.5) } is an
orthonormal basis for the hyperplane.  Let (a,b,c) be a point on the
hyperplane with respect to this basis.  (a,b,c) is in the 4-cube if and
only if |a| + |b| + |c| <= 2.   The shape of the intersection is a
regular octahedron.

==> geometry/tetrahedron.p <==
Suppose you have a sphere of radius R and you have four planes that are
all tangent to the sphere such that they form an arbitrary tetrahedron
(it can be irregular).  What is the ratio of the surface area of the
tetrahedron to its volume?

==> geometry/tetrahedron.s <==
For each face of the tetrahedron, construct a new tetrahedron with that
face as the base and the center of the sphere as the fourth vertex.
Now the original tetrahedron is divided into four smaller ones, each of
height R.  The volume of a tetrahedron is Ah/3 where A is the area of
the base and h the height; in this case h=R.  Combine the four
tetrahedra algebraically to find that the volume of the original
tetrahedron is R/3 times its surface area.

==> geometry/tiling/rational.sides.p <==
A rectangular region R is divided into rectangular areas.  Show that if
each of the rectangles in the region has at least one side with
rational length then the same can be said of R.

==> geometry/tiling/rational.sides.s <==
"Fourteen proofs of a result about tiling a rectangle" (Stan Wagon)
_The American Mathematical Monthly_, Aug-Sep 1987, Vol 94 #7.  There
was also a fifteenth proof published a few issues later, attributed to
a (University of Kentucky?) student.

==> geometry/tiling/rectangles.with.squares.p <==
Given two sorts of squares, (axa) and (bxb), what rectangles can be tiled?

==> geometry/tiling/rectangles.with.squares.s <==
A rectangle can be tiled with (axa) and (bxb) squares,   iff

(i) gcd(a,b)=1 , and any of the following hold:

either:  both sides of the rectangle are multiples of a;
    or:  both sides of the rectangle are multiples of b;
    or:  one side is a multiple of (ab), and the other is any length EXCEPT
         one of a finite number of "bad" lengths: those numbers which are
         NOT positive integer combinations of a & b. { By Sylvester's theorem
         there are (a-1)(b-1)/2 of these, the largest being (a-1)(b-1)-1. }

(ii) gcd(a,b) = d .
     Then merely apply (i) to the problem with a,b replaced
     by a/d, b/d  and the rectangle lengths also divided by d.
     i.e.  all cells must appear in (dxd) subsquares.

------
PROOF
It is clear that (ii) follows from (i), and that simple constructions give
the "if" part of (i). For the "only if" part, we prove that...

(S) If one side of the rectangle is not divisible by a, and the other is
    not divisible by b, then the tiling is impossible.

The results in (i) follow immediately from (S).

To prove (S):  ( Chakraborty-Hoey style ).
                 ~~~~~~~~~~~~~~~~
Let the width of the rectangle be a NON-(a-multiple). Then the number of
bxb squares starting (i.e. top edge) at row 1 must be a NON-a-multiple.
Thus the number of bxb starting at row 2 must BE an a-multiple. Similarly
for the number starting at rows 3,4,...,b . Then the number starting at
row (b+1) must be a NON-a-multiple again.

Similarly the number starting at rows (2b+1), (3b+1), (4b+1),... must all be
non-a-multiples. So if the number of rows is NOT a multiple of b, (call it
bx+r), then row (bx+1) must have a NON-a-multiple of bxb squares starting
there, i.e. at least one, and there is no room left to squeeze it in.     [QED]
----

A Rickard-style proof of (S) is    ..BBB....BBWWW...WBBB....BBWWW...W(..etc)
  ~~~~~~~   also possible, by      ..BBB....BBWWW...WBBB....BBWWW...W
coloring the rectangle in          ..BBB....BBWWW...WBBB....BBWWW...W
vertical strips as shown here:       <-  a  ->< b-a ><-  a  ->< b-a >

Every square tile covers an a-multiple of black squares. But if the width
is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there
are a NON-a-multiple of black squares in total.    [QED]

(Note: the coloring must have 1 column of blacks on the right, and any
 ====     spare columns of whites on the left.)

===================

Bill Taylor.            [e-mail]

>A Rickard-style proof of (S) is    ..BBB....BBWWW...WBBB....BBWWW...W(..etc)
>  ~~~~~~~   also possible, by      ..BBB....BBWWW...WBBB....BBWWW...W
>coloring the rectangle in          ..BBB....BBWWW...WBBB....BBWWW...W
>vertical strips as shown here:       <-  a  ->< b-a ><-  a  ->< b-a >
>
>Every square tile covers an a-multiple of black squares. But if the width
>is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there
>are a NON-a-multiple of black squares in total.    [QED]
>
>(Note: the coloring must have 1 column of blacks on the right, and any
> ====     spare columns of whites on the left.)

This statement of how to position the colouring isn't good enough, I'm
afraid. Take a=4, b=7 and consider e.g. a 19x10 rectangle. Coloured your
way, you get:

    BWWWBBBBWWWBBBBWWWB
    BWWWBBBBWWWBBBBWWWB
    :::::::::::::::::::
    BWWWBBBBWWWBBBBWWWB
    BWWWBBBBWWWBBBBWWWB

The result has 10*10=100 black squares in it, which *is* a multiple of a=4,
despite the fact that 19 is not a multiple of 7 and 10 is not a multiple of
4.

Of course, there is an alternative offset for the pattern that does give you
the result you want:

    WWBBBBWWWBBBBWWWBBB
    WWBBBBWWWBBBBWWWBBB
    :::::::::::::::::::
    WWBBBBWWWBBBBWWWBBB
    WWBBBBWWWBBBBWWWBBB

To show this happens in general: because the width of the rectangle is a
non-multiple of b, it is possible to position it on the pattern so that the
leftmost column in the rectangle is white and the column just right of the
right edge of the rectangle is black. Suppose N columns are black with this
positioning. Then the rectangle contains N*H black cells, where H is the
height of the rectangle.

If we then shift the rectangle right by one, the number of black columns
increases by 1 and it contains (N+1)*H black cells. The difference between
these two numbers of black cells is H, which is not a multiple of a.
Therefore N*H and (N+1)*H cannot both be multiples of a, and so one of these
two positionings of the pattern will suit your purposes.

David Seal
[e-mail]

==> geometry/tiling/scaling.p <==
A given rectangle can be entirely covered (i.e. concealed) by an
appropriate arrangement of 25 disks of unit radius.

Can the same rectangle be covered by 100 disks of 1/2 unit radius?

==> geometry/tiling/scaling.s <==
Yes.  The same configuration of circles, when every distance is reduced
by half (including the diameters), will cover a similar rectangle whose
sides are one half of the original one.  The original rectangle is the
union of four such rectangles.

==> geometry/tiling/seven.cubes.p <==
Consider 7 cubes of equal size arranged as follows. Place 5 cubes so
that they form a Swiss cross or a + (plus). ( 4 cubes on the sides and
1 in the middle). Now place one cube on top of the middle cube and the
seventh below the middle cube, to effectively form a 3-dimensional
swiss cross.

Can a number of such blocks (of 7 cubes each) be arranged so that they
are able to completely fill up a big cube (say 10 times the size of
the small cubes)? It is all right if these blocks project out of the
big cube, but there should be no holes or gaps.

==> geometry/tiling/seven.cubes.s <==
Let n be a positive integer.  Define the function f from Z^n to Z by
f(x) = x_1+2x_2+3x_3+...+nx_n.  For x in Z^n, say y is a neighbor of x
if y and x differ by one in exactly one coordinate.  Let S(x) be the
set consisting of x and its 2n neighbors.  It is easy to check that
the values of f(y) for y in S(x) are congruent to 0,1,2,...,2n+1 (mod
2n+1) in some order.  Using this, it is easy to check that every y in
Z^n is a neighbor of one and only one x in Z^n such that f(x) is
congruent to 0 (mod 2n+1).  So Z^n can be tiled by clusters of the
form S(x), where f(x) is congruent to 0 mod 2n+1.

==> group/group.01.p <==
AEFHIKLMNTVWXYZ BCDGJOPQRSU

==> group/group.01.s <==
AEFHIKLMNTVWXYZ    drawn with straight lines
BCDGJOPQRSU    not drawn with straight lines

==> group/group.01a.p <==
147 0235689

==> group/group.01a.s <==
147    drawn with straight lines
0235689    not drawn with straight lines

==> group/group.02.p <==
ABEHIKMNOPTXZ CDFGJLQRSUVWY

==> group/group.02.s <==
ABEHIKMNOPTXZ    resembles Greek letter
CDFGJLQRSUVWY    does not resemble Greek letter

==> group/group.03.p <==
BEJQXYZ DFGHLPRU KSTV CO AIW MN

==> group/group.03.s <==

BEJQXYZ            no state starting with this letter
DFGHLPRU        one state starting with this letter
KSTV            two states starting with this letter
CO            three states starting with this letter
AIW            four states starting with this letter
            five states starting with this letter
            six states starting with this letter
            seven states starting with this letter
MN            eight states starting with this letter

==> group/group.04.p <==
BDO P ACGJLMNQRSUVWZ EFTY HIKX

==> group/group.04.s <==
BDO        no endpoint
P        one endpoint
ACGJLMNQRSUVWZ    two endpoints
EFTY        three endpoints
HIKX        four endpoints

==> group/group.05.p <==
CEFGHIJKLMNSTUVWXYZ ADOPQR B

==> group/group.05.s <==
CEFGHIJKLMNSTUVWXYZ    no enclosed area
ADOPQR            one enclosed area
B            two enclosed areas

==> group/group.06.p <==
BCEGKMQSW DFHIJLNOPRTUVXYZ

==> group/group.06.s <==
BCEGKMQSW        prime numbers
DFHIJLNOPRTUVXYZ    composites

Xref: bloom-picayune.mit.edu rec.puzzles:18146 news.answers:3077
Newsgroups: rec.puzzles,news.answers
Path: bloom-picayune.mit.edu!enterpoop.mit.edu!snorkelwacker.mit.edu!usc!wupost!spool.mu.edu!uunet!questrel!chris
From: uunet!questrel!chris (Chris Cole)
Subject: rec.puzzles FAQ, part 11 of 15
Message-ID: <[e-mail]>
Followup-To: rec.puzzles
Summary: This posting contains a list of
     Frequently Asked Questions (and their answers).
     It should be read by anyone who wishes to
     post to the rec.puzzles newsgroup.
Sender: [e-mail] (Chris Cole)
Reply-To: uunet!questrel!faql-comment
Organization: Questrel, Inc.
References: <[e-mail]>
Date: Mon, 21 Sep 1992 00:09:38 GMT
Approved: [e-mail]
Expires: Sat, 3 Apr 1993 00:08:21 GMT
Lines: 1889

Archive-name: puzzles-faq/part11
Last-modified: 1992/09/20
Version: 3

==> induction/hanoi.p <==
Is there an algorithom for solving the hanoi tower puzzle for any number
of towers?  Is there an equation for determining the minimum number of
moves required to solve it, given a variable number of disks and towers?

==> induction/hanoi.s <==
The best way of thinking of the Towers of Hanoi problem is inductively.
To move n disks from post 1 to post 2, first move (n-1) disks
from post 1 to post 3, then move disk n from post 1 to post 2, then move
(n-1) disks from post 3 to post 2 (same procedure as moving (n-1) disks
from post 1 to post 3).  In order to figure out how to move (n-1) disks
from post 1 to post 3, first move (n-2) disks . . . .

As far as an algorithm which straightens out any legal position is
concerned, the algorithm would go something like this:

1.  Find the smallest disk.  Call the post that it's on post 1.

2.  Find the smallest disk which is not on post 1.  This disk is, say,
size k.  (I am calling the smallest disk size 1 here.)  Call the post
that disk k is on post 2.  Disks 1 through (k-1) are then all stacked up
correctly on post 1 disk k is on top of post 2.  This follow from the
fact that the disks are in a legal postition.

3.  Move disks 1 through (k-1) from post 1 to post 2, ignoring the other
disks.  This is just the standard Tower of Hanoi problem for (k-1)
disks.

4.  If the disks are not yet correctly arranged, repeat from step 1.

In fact, this gives the straightening with the fewest number of moves.

With 3 towers, for N number of disks, the formula for the minimum number of
moves to complete the puzzle correctly is:
        (2^N) - 1

This bit of ancient folklore was invented by De Parville in 1884.

``In the great temple at Benares, says he, beneath the dome which
  marks the centre of the world, rests a brass plate in which are
  fixed three diamond needles, each a cubit high and as thick as
  the body of a bee.  On one of these needles, at the creation,
  God placed sixty-four discs of pure gold, the largest disc resting
  on the brass plate, and the others getting smaller and smaller
  up to the top one.  This is the Tower of Bramah.  Day and night
  unceasingly the priests transfer the discs from one diamond needle
  to another according to the fixed and immutable laws of Bramah,
  which require that the priest on duty must not move more than one
  disc at a time and that he must place this disc on a needle so
  that there is no smaller disc below it.  When the sixty-four
  discs shall have been thus transferred from the needle on which
  at the creation God placed them to one of the other needles,
  tower, temple, and Brahmins alike will crumble into dust, and
  with a thunderclap the world will vanish.''  (W W R Ball,
  MATHEMATICAL RECREATIONS AND ESSAYS, p. 304)

This has been discussed by several authors, e.g.

    Er, Info Sci 42 (1987) 137-141.
    Graham, Knuth and Patashnik, _Concrete_Mathematics_.

There are many papers claiming to solve this, and they are probably
all correct but they rely on the unproven "Frame's conjecture".
In particular, for the 4 peg case the conjecture states that an optimal
solution begins by forming a substack of the k smallest discs, then moving
the rest, and then moving those k again; k to be determined.

Here is a extensible bc program that does the same work. The output
format is not that great. We get 300 numbers as output. The first
hundred represent N, the next 100 represent f(N) and the last hundred
represent i, which is the number of discs to move to tmp1 using f(N).
For convenience, I have here some values for N <= 48. Enjoy.

Sharma


N     1   2   3  4  5   6  7  8  9  10  11 12 13  14  15  16  17  18  19
f(N)     1   3   5  9 13  17 25 33 41  49  65 81 97 113 129 161 193 225 257
i     0   1   1  2  2   3  3  4  5   6   6  7  8   9  10  10  11  12  13


N    20   21  22  23  24  25  26  27  28  29  30   31   32   33   34   35
f(N)    289 321 385 449 513 577 641 705 769 897 1025 1153 1281 1409 1537 1665
i    14   15  15  16  17  18  19  20  21  21   22   23   24   25   26   27

N    36    37    38    39     40    41   42   43   44   45   46   47   48
f(N)    1793 2049  2305  2561   2817  3073 3329 3585 3841 4097 4609 5121 5633
i    28    28    29    30     31    32    33   34   35   36  36   37   38


/* This is the bc program that gives f(N) for 4 peg case */

w = 101; /* This represents the number of disks */

m[0] = 0;
m[1] = 1;
m[2] = 3;
m[3] = 5;
m[4] = 9;
m[5] = 13;
m[6] = 17;

/* f(n) is the function that gives the min # of moves for 4 peg case */
define f(n) {
  return (m[n]);
}

/* g(n) is the function that fives the min # of moves for 3 peg case */
define g(n) {
  return (2^n - 1);
}

/* x(n) is the Optimization Routine */

define x(n) {
  auto j
  auto r
  auto i

  if(n == 1) return (1);
  j = f(1) + g(n-1);

  for(i = 2; i < n; i++) {
    r = f(i) + g(n-i);
    if(r < j) { j = r; d = i; }
  }
  return (j);
}

/* main program */
for(q = 4; q < w; q++) {
  t = x(q-1);
  m[q] = 2 * t + 1;
  d[q] = d;
};


/*This for loop prints the number of discs from 1 <= n <=  w*/

for(q = 1; q < w; q++) {
q;
}

/*This for loop prints f(n) for 1 <= n <= w */

for(q = 1; q < w; q++) {
m[q];
}

/*This for loop prints i for 1 <= n <= w
i represents the number of disks to be moved to tmp location using f(n)
N-i-1 will be moved using g(n) */

for(q = 1; q < w; q++) {
d[q];
}

--
[e-mail]

==> induction/n-sphere.p <==
With what odds do three random points on an n-sphere form an acute triangle?

==> induction/n-sphere.s <==
Select three points a, b, and c, randomly with respect to the surface of an
n-sphere.  These three points determine a fourth, x, which is the intersection
of the sphere with the axis perpendicular to the abc plane.  (Choose the pole
nearest the plane.) I could have, just as easily, selected x, a distance d
from x, and three points d units away from x.  The distribution of d is not
uniform, but that's ok.  For every x and d, the three points abc form an acute
triangle with probability p[n-1].  By induction, p[n] = 1/4.

==> induction/paradox.p <==
What simple property holds for the first 10,000 integers, then fails?

==> induction/paradox.s <==
Consider the sequences defined by:
s(1) = a; s(2) = b; s(n) = least integer such that s(n)/s(n-1) > s(n-1)/s(n-2).
In other words, s(n) = 1+floor(s(n-1)^2/s(n-2)) for n >= 3.  These
sequences are similar in some ways to the classically-studied Pisot
sequences.  For example, if a = 1, b = 2, then we get the odd-indexed
Fibonacci numbers.

D. Boyd of UBC, an expert in Pisot sequences, pointed out the following.
If we let a = 8, b = 55 in the definition above, then the resulting
sequence s(n) appears to satisfy the following linear recurrence
of order 4:

    s(n) = 6s(n-1) + 7s(n-2) - 5s(n-3) - 6s(n-4)

Indeed, it does satisfy this linear recurrence for the
first 11,056 terms.  However, it fails at the 11,057th term!
And s(11057) is a 9270 digit number.
(The reason for this coincidence depends on a remarkable fact
about the absolute values of the roots of the polynomial
x^4 - 6x^3 - 7x^2 + 5x + 6.)

==> induction/party.p <==
You're at a party.  Any two (different) people at the party have exactly one
friend in common (the friend is also at the party).  Prove that there is at
least one person at the party who is a friend of everyone else.  Assume that
the friendship relation is symmetric and not reflexive.

==> induction/party.s <==
Here is an easy solution by induction. Let P be the set of people in the
party, and n the size of P. If n=2 or 3, the result is trivial. Suppose now
that n>3 and that the result is true for n-1.

For any two distinct x,y in P, write x & y to mean that `x is a friend of y',
and x ~& y to mean that `x is not a friend of y'.

Take q in P. The hypothesis on the relation & is still satisfied on P-{q}; by
induction, the result is thus true for P-{q}, and there is some p in P-{q}
such that p & x for any x in P-{p,q}. We have two cases:

a) p & q. Then the result holds for P with p.

b) p ~& q. By hypothesis, there is a unique r in P-{p,q} such that p & r & q.
For any x in P-{p,q}, if x & q, then p & x & q, and so x=r. Thus r is the
unique friend of q. Now for any s in P-{q,r} there exists some x such that s &
x & q, and so x=r. This means that r & s for any s in P-{q,r}, and as r & q,
the results holds in P with r.

The problem can also be solved by applying the spectral theory of graphs
(see for instance Bollobas' excellent book, _Extremal Graph Theory_).

The problem's condition is vacuous if there is only N=1 person at the "party",
impossible if N=2 (If you aren't your own friend, nor I mine, somebody *else*
must be our mutual friend), and trivial if N=3 (everybody must be everyone
else's friend).  Henceforth assume N>3.

Let A,B be two friends, and C their mutual friend.  Let a be the number
of A's friends other than B and C, and likewise b,c.  Each of A's friends
is also friendly with exactly one other of A's friends, and with none of
B and C's other friends (if A1,B1 are friends of A,B resp. and of each other
then A1 and B have more than one mutual friend); likewise for B and C.
Let M=N-(a+b+c+3) be the number of people not friendly with any of A,B,C.
Each of them is friendly with exactly one of A's and one of B's friends;
and each pair of a friend of A and a friend of B must have exactly
one of them as a mutual friend.  Thus M=ab; likewise M=ab=ac=bc. Thus
either M and two of a,b,c vanish, or a=b=c=k (say), M=k^2, and N=k^3+3k+3.
In the first case, say b=c=0; necessarily a is even, and A is a friend of
everybody else at the party, each of whom is friendly with exactly one other
person; clearly any such configuration (a graph of k/2+1 triangles with a
common vertex) satisfies the problem's conditions).

It remains to show that the second case is impossible.  Since N=k^2+3k+3
does not depend on A,B,C, neither does k, and it quickly follows that the
party's friendship graph is regular with reduced matrix

         [  0   k+2   0  ]
         [  1    1    k  ]
         [  0    1   k+1 ]

and eigenvalues k+2 and +-sqrt(k+1) and multiplicities 1,m1,m2 for some
*integers* m1 and m2 such that (m1-m2)*sqrt(k+1)=-(k+2) (because the graph's
matrix has trace zero).  Thus sqrt(k+1) divides k+2 and k+1 divides

        (k+2)^2=(k+1)(k+3)+1

which is only possible if k=0,  Q.E.D.

==> induction/roll.p <==
An ordinary die is thrown until the running total of the throws first
exceeds 12.  What is the most likely final total that will be obtained?

==> induction/roll.s <==
Claim: If you throw a die until the running total exceeds n>=5, a final
outcome of n+1 is more likely than any other.

Assume we throw an m for a total n+k>n+1, and assume m-k>=0.  Now, it
is just as likely to throw an m as an m-k+1, which means that the sum
n+1 is just as likely as any other.  Now consider the series of throws
consisting of n-5 1's followed by a 6 and note that we cannot achieve
more than an n+1 by changing the last die roll.  Hence, a total of n+1
is more likely than any other.

==> induction/takeover.p <==
After graduating from college, you have taken an important managing position
in the prestigious financial firm of "Mary and Lee".
You are responsable for all the decisions concerning take-over bids.
Your immediate concern is whether to take over "Financial Data".
There is no doubt that you will be successful if you are the first to
bid and that this will be profitable for the firm and you in the long
run.  However, you know that there exist another n financial firms,
similar to "Mary and Lee", that are also considering the possibility.
Although you are likely to be the first one to move, you know that
just after a take-over there is a lot of adjustment that needs to be
done.  In fact, for a period of time following any take-over the
successful firm becomes a prime candidate for a take-over which will
cost the job of whoever is responsable for take-overs.  Among all
financial firms it is common knowledge that the managers responsable
for take-overs are rational and intelligent.  What is your best response?

==> induction/takeover.s <==
Assume the takeover is wise for n.  The takeover is then unwise for
n+1, as the other companies now find themselves in the same situation
as you for n.  If the decision is unwise for n, by similar reasoning
it is wise to takeover FD for n+1.  Now note that for n=1 the takeover
decision is clearly unwise, hence by induction you should takeover
FD iff n is even.

==> logic/29.p <==
Three people check into a hotel.  They pay $30 to the manager and go
to their room.  The manager finds out that the room rate is $25 and
gives $5 to the bellboy to return.  On the way to the room the bellboy
reasons that $5 would be difficult to share among three people so
he pockets $2 and gives $1 to each person.

Now each person paid $10 and got back $1.  So they paid $9 each,
totalling $27.  The bellboy has $2, totalling $29.

Where is the remaining dollar?

==> logic/29.s <==
Each person paid $9, totalling $27.  The manager has $25 and the bellboy $2.
The bellboy's $2 should be added to the manager's $25 or subtracted from
the tenants' $27, not added to the tenants' $27.

==> logic/ages.p <==
1) Ten years from now Tim will be twice as old as Jane was when Mary was
   nine times as old as Tim.

2) Eight years ago, Mary was half as old as Jane will be when Jane is one year
   older than Tim will be at the time when Mary will be five times as old as
   Tim will be two years from now.